Chapter 7 - Analytic Trigonometry - 7.2 The Inverse Trigonometric Functions (Continued) - 7.2 Assess Your Understanding - Page 457: 10

$\dfrac{\sqrt3}{2}$

Note that $\cos^{-1}{\left(\dfrac{1}{2}\right)}=\frac{\pi}{3}$, because $\cos{\left(\dfrac{\pi}{3}\right)}=\frac{1}{2}$ and $\frac{\pi}{3}$ is in the range of $\cos^{-1}{x}$, which is $\left[0,\pi\right].$ Thus $\sin{\left(\cos^{-1}{\left(\dfrac{1}{2}\right)}\right)}=\sin{\dfrac{\pi}{3}}=\dfrac{\sqrt3}{2}$